Question

when the displacement is half of the amplitude then what fraction of total energy of simple harmonic oscillator is kinetic

Answer 1

total energy = kinetic energy = potential energy

therefore t.e=1/2k(a^2-x^2)+1/2kx^2

, since x =1/2a

the kinetic energy is equal to 1/2k(a^2--1/4a^2)

on solving the above the kinetic energy will become equal to

3/8ka^2

Answer 2

Answer: The fraction of total and kinetic energy will be $\dfrac{T.E}{K.E}=\dfrac{4}{3}$.

Explanation:

Given that,

Displacement $x =\dfrac{a}{2}$

We know that,

The total energy of simple harmonic oscillator is

$T.E =K.E+P.E$

$E = \dfrac{1}{2}k(a^2-x^2)+\dfrac{1}{2}kx^2$

$E = \dfrac{1}{2}ka^2$

The potential energy is

$P.E = \dfrac{1}{2}kx^2$

Put the value of x

$P.E = \dfrac{1}{2}k(\dfrac{a}{2})^2$

$P.E = \dfrac{1}{8}ka^2$

$P.E = \dfrac{1}{4}(\dfrac{1}{2}ka^2)$

$P.E= \dfrac{1}{4}E$

The kinetic energy is

$K.E = T.E-P.E$

$K.E = E-\dfrac{1}{4}E$

$K.E = \dfrac{3E}{4}$

The kinetic energy is $\dfrac{3}{4}$ of total energy.

The fraction of total and kinetic energy is

$\dfrac{T.E}{K.E}=\dfrac{E4}{3E}$

$\dfrac{T.E}{K.E}=\dfrac{4}{3}$

Hence, The fraction of total and kinetic energy will be $\dfrac{T.E}{K.E}=\dfrac{4}{3}$.